Bott–Samelson atlases, total positivity, and Poisson structures on some homogeneous spaces

Abstract

Let G be a connected and simply connected complex semisimple Lie group. For a collection of homogeneous G-spaces G/Q, we construct a finite atlas ${{\mathcal {A}}}_{{\scriptscriptstyle BS}}(G/Q)$ on G/Q, called the Bott–Samelson atlas, and we prove that all of its coordinate functions are positive with respect to the Lusztig positive structure on G/Q. We also show that the standard Poisson structure $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}$ on G/Q is presented, in each of the coordinate charts of ${{\mathcal {A}}}_{{\scriptscriptstyle BS}}(G/Q)$, as a symmetric Poisson CGL extension (or a certain localization thereof) in the sense of Goodearl–Yakimov, making $(G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}, {{\mathcal {A}}}_{{\scriptscriptstyle BS}}(G/Q))$ into a Poisson–Ore variety. In addition, all coordinate functions in the Bott–Samelson atlas are shown to have complete Hamiltonian flows with respect to the Poisson structure $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}$. Examples of G/Q include G itself, G/T, G/B, and G/N, where $T \subset G$ is a maximal torus, $B \subset G$ a Borel subgroup, and N the uniradical of B.

Appendix A. Proof of Theorem 5.3 and T-leaves of \((G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})\)

In this appendix, we first prove Theorem 5.3 which says that for any $v, w \in W$,

$$\begin{aligned} J^w_{{\scriptscriptstyle B}(v)}: \;\;&\; (wB^-B/B(v), \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)}) \longrightarrow \left( {{\mathcal {O}}}^{w_0w^{-1}} \times {{\mathcal {O}}}^{(w, v)}, \; \pi _{1, 2}\right) , \end{aligned}$$

(A.1)

$$\begin{aligned} J^w_{{\scriptscriptstyle N}(v)}: \;\;&\; (wB^-B/N(v), \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) \longrightarrow \left( {{\mathcal {O}}}^{w_0w^{-1}} \times {{\mathcal {O}}}^{(w, v)} \times T, \; \pi _{1, 2, 0}\right) \end{aligned}$$

(A.2)

are Poisson isomorphisms, where $J^w_{{\scriptscriptstyle B}(v)}$ and $J^w_{{\scriptscriptstyle N}(v)}$ are given in (2.25) and (2.26), and the Poisson structures $\pi _{1, 2}$ and $\pi _{1, 2,0}$ are given in (5.2) and (5.3). Here recall that for $Q = B(v)$ or N(v), $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}$ is the projection to G/Q of the standard Poisson structure $\pi _{\mathrm{st}}$ on G. In Sect. A.1, we review some facts on the Poisson Lie group $(G, \pi _{\mathrm{st}})$. In Sect. A.2, we prove certain maps involved in the definitions of $J^w_{{\scriptscriptstyle B}(v)}$ and $J^w_{{\scriptscriptstyle N}(v)}$ are Poisson, and we use these facts to prove Theorem 5.3 in Sect. A.3. In Sect. A.4, we determine the T-leaves of $(G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})$.

1.1 Some facts on the Poisson Lie group $(G, \pi _{\mathrm{st}})$

We first recall (see, for example, [16, 39]) that given a Poisson Lie group $(L, \pi )$ and a Poisson manifold $(X, {\pi _{\scriptscriptstyle X}})$, a left action of L on X is said to be Poisson if the action map $(L, \pi ) \times (X, {\pi _{\scriptscriptstyle X}}) \rightarrow (X, {\pi _{\scriptscriptstyle X}})$ is Poisson. A Poisson manifold $(X, {\pi _{\scriptscriptstyle X}})$ is called a Poisson homogeneous space [14] of a Poisson Lie group $(L, \pi )$ if $(X, {\pi _{\scriptscriptstyle X}})$ has a Poisson action by $(L, \pi )$ which is also transitive.

Example A.1

If M is a closed coisotropic subgroup (see Sect. 5.1) of a Poisson Lie group $(L, \pi )$, the action of L on L/M by left translation makes $(L/M, \pi _{{\scriptscriptstyle L}/{\scriptscriptstyle M}})$ a Poisson homogeneous space of $(L, \pi )$, where $\pi _{{\scriptscriptstyle L}/{\scriptscriptstyle M}}$ is the projection to L/M of the Poisson structure $\pi$ on L. Note that as $\pi (e) = 0$, $\pi _{{\scriptscriptstyle L}/{\scriptscriptstyle M}}$ vanishes at $e_\cdot M \in L/M$. In general, it is easy to see from the definitions that if $(X, {\pi _{\scriptscriptstyle X}})$ is a Poisson homogeneous space of $(L, \pi )$ and if $x \in X$ is such that ${\pi _{\scriptscriptstyle X}}(x)=0$, then the stabilizer subgroup $L_x$ of L at x is a coisotropic subgroup of $(L, \pi )$, and the map

$$\begin{aligned} (L/L_x, \pi _{{\scriptscriptstyle L}/{{\scriptscriptstyle L}}_x}) \longrightarrow (X, {\pi _{\scriptscriptstyle X}})\;\; l \longmapsto lx, \quad l \in L, \end{aligned}$$

is a Poisson isomorphism. $\square$

Returning to the Poisson Lie group $(G, \pi _{\mathrm{st}})$, where $\pi _{\mathrm{st}}$ is given in (4.9), for any $v \in W$ and $Q = B(v)$ or N(v), the Poisson manifold $(G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})$ is then a Poisson homogeneous space of $(G, \pi _{\mathrm{st}})$.

We now recall a Drinfeld double of the Poisson Lie group $(G, \pi _{\mathrm{st}})$: associated to the standard quasi-triangular r-matrix $r_{\mathrm{st}} \in {\mathfrak {g}}\otimes {\mathfrak {g}}$ in (4.8), one has the quasi-triangular r-matrix $r_{\mathrm{st}}^{(2)}\in ({\mathfrak {g}}\oplus {\mathfrak {g}})^{\otimes 2}$ for the direct product Lie algebra ${\mathfrak {g}}\oplus {\mathfrak {g}}$, given [39, Sect. 6.1] as (see Notation 4.13)

$$\begin{aligned} r_{\mathrm{st}}^{(2)} = (r_{\mathrm{st}}, \; 0) -(0, \, r_{\mathrm{st}}^{21}) - \sum _{q=1}^d (H_q, 0) \wedge (0, H_q) -\sum _{\alpha \in \Delta ^+} \langle \alpha , \alpha \rangle (e_\alpha , 0) \wedge (0, e_{-\alpha }),\qquad \end{aligned}$$

(A.3)

where $r_{\mathrm{st}}^{21} = \tau (r_{\mathrm{st}})$ with $\tau (x \otimes y) = y \otimes x$ for $x, y \in {\mathfrak {g}}$, and for any vector space V and $u, v \in V$, we use the convention that

$$\begin{aligned} u \wedge v = u\otimes v - v\otimes u \in V \otimes V. \end{aligned}$$

Let $\Lambda _{\mathrm{st}}^{(2)} \in \wedge ^2 ({\mathfrak {g}}\oplus {\mathfrak {g}})$ be the skew-symmetric part of $r_{\mathrm{st}}^{(2)}$, and let $\Pi _{\mathrm{st}}$ be the multiplicative Poisson structure on the product group $G \times G$ given by

$$\begin{aligned} \Pi _{\mathrm{st}}= \left( r_{\mathrm{st}}^{(2)}\right) ^L - \left( r_{\mathrm{st}}^{(2)}\right) ^R= \left( \Lambda _{\mathrm{st}}^{(2)}\right) ^L- \left( \Lambda _{\mathrm{st}}^{(2)}\right) ^R. \end{aligned}$$

(A.4)

Here, for $A \in {\mathfrak {g}}^{\otimes k}$, $A^L$ and $A^R$ respectively denote the left and right invariant tensor fields on G with value A at the identity of G. It follows from the definitions that

$$\begin{aligned} \Pi _{\mathrm{st}}= (\pi _{\mathrm{st}}, \, 0) + (0, \, \pi _{\mathrm{st}}) + \mu _1 + \mu _2, \end{aligned}$$

(A.5)

where

$$\begin{aligned} \mu _1&= \sum _{q=1}^d (H_q^R, \; 0) \wedge (0, \; H_q^R) +\sum _{\alpha \in \Delta ^+} \langle \alpha , \alpha \rangle (e_\alpha ^R, \, 0) \wedge (0, e_{-\alpha }^R), \end{aligned}$$

(A.6)

$$\begin{aligned} \mu _2&=-\sum _{q=1}^d (H_q^L, \; 0) \wedge (0, \; H_q^L) -\sum _{\alpha \in \Delta ^+} \langle \alpha , \alpha \rangle (e_\alpha ^L, \, 0) \wedge (0, e_{-\alpha }^L). \end{aligned}$$

(A.7)

The Poisson Lie group $(G \times G, \, \Pi _{\mathrm{st}})$ is a Drinfeld double of the Poisson Lie group $(G, \pi _{\mathrm{st}})$ (see, for example, [39, paragraph after Example 6.11]). In particular, the embedding

$$\begin{aligned} (G, \, \pi _{\mathrm{st}}) \hookrightarrow (G \times G, \; \Pi _{\mathrm{st}}), \;\; g \longmapsto (g, g), \quad g \in G, \end{aligned}$$

is Poisson, and the projections $(G \times G, \Pi _{\mathrm{st}}) \rightarrow (G, \pi _{\mathrm{st}})$ to both factors are Poisson.

It follows from (A.5), (A.6) and (A.7) that $B \times B$ is a coisotropic subgroup of the Poisson Lie group $(G \times G, \, \Pi _{\mathrm{st}})$. Let $\varpi$ be the projection

$$\begin{aligned} \varpi : \;\; G \times G \longrightarrow (G \times G)/(B \times B) \cong G/B \times G/B, \end{aligned}$$

and let $\Pi = \varpi (\Pi _{\mathrm{st}})$. It follows from (A.4) and (A.5) that

$$\begin{aligned} \Pi =- \varpi \left( \left( \Lambda _{\mathrm{st}}^{(2)}\right) ^R\right) = (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}, \, 0) + (0, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})+ \varpi (\mu _1). \end{aligned}$$

Let $G_{\mathrm{diag}} = \{(g, g): g \in G\} \subset G \times G$ and consider the $G_{\mathrm{diag}}$-orbits in $G/B \times G/B$, which are precisely of the form

$$\begin{aligned} G_{\mathrm{diag}} (v) \; {\mathop {=}\limits ^{\mathrm{def}}}\; G_{\mathrm{diag}}(e_\cdot B, \, {\overline{v}}_\cdot B)\subset G/B \times G/B, \quad v \in W. \end{aligned}$$

Note that for $v \in W$, the stabilizer subgroup of $G \cong G_{\mathrm{diag}}$ at $(e_\cdot B, {\overline{v}}_\cdot B) \in G/B \times G/B$ is precisely $B(v) = B \cap {\overline{v}}B {\overline{v}}^{\, -1}$.

Lemma A.2

For each $v \in W$, $G_{\mathrm{diag}} (v)$ is a Poisson submanifold of $G/B \times G/B$ with respect to $\Pi$, and the G-equivariant map

$$\begin{aligned} (G/B(v), \; \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}}(v)}) \longrightarrow (G_{\mathrm{diag}} (v), \, \Pi ),\;\; g_\cdot B(v) \longmapsto (g_\cdot B, \; g{\overline{v}}_\cdot B), \quad g \in G,\qquad \end{aligned}$$

(A.8)

is a Poisson isomorphism.

Proof

Let $v \in W$. By [42, Theorem 2.3], $G_{\mathrm{diag}} (v)$ is a Poisson submanifold of $G/B \times G/B$ with respect to $\Pi$, and, as a $G \cong G_{\mathrm{diag}}$-orbit, $(G_{\mathrm{diag}} (v), \Pi )$ is a Poisson homogeneous space of $(G, \pi _{\mathrm{st}})$. It is also easy to see that $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}({\overline{v}}_\cdot B) = 0$ and $\varpi (\mu _1)(e_\cdot B, {\overline{v}}_\cdot B) = 0$. It follows that $\Pi (e_\cdot B, {\overline{v}}_\cdot B) = 0$. By Example A.1, the map in (A.8) is an isomorphism of Poisson homogeneous spaces of $(G, \pi _{\mathrm{st}})$. $\square$

Recall the Poisson manifold $(F_2, \pi _2)$ from Sect. 4.4. By [39, Theorem 7.8] (see also [40, Proposition 5.6]), the map

$$\begin{aligned} J_2: \;\; (F_2, \, \pi _2) \longrightarrow (G/B \times G/B, \; \Pi ), \;\; [g_1, g_2]_{F_2} \longmapsto ({g_1}_\cdot B, \, {g_1g_2}_\cdot B), \end{aligned}$$

(A.9)

is a Poisson isomorphism, with

$$\begin{aligned} J_2^{-1}: \;\; (G/B \times G/B, \; \Pi ) \longrightarrow (F_2, \pi _2), \;\; ({h_1}_\cdot B, \, {h_2}_\cdot B) \longmapsto [h_1, \, h_1^{-1}h_2]_{{\scriptscriptstyle F}_2}.\nonumber \\ \end{aligned}$$

(A.10)

Note that $J_2$ is G-equivariant if $F_2$ is given the G-action by

$$\begin{aligned} g_\cdot [g_1, \, g_2]_{F_2} = [gg_1, \; g_2]_{F_2}, \quad g, g_1, g_2 \in G. \end{aligned}$$

(A.11)

By Lemma A.2, we have the following interpretation of the Poisson homogeneous space $(G/B(v), \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}(v)}})$ as a Poisson submanifold in $(F_2, \pi _2)$.

Lemma A.3

For any $v \in W$, the map

$$\begin{aligned} E_v:\;\; (G/B(v), \; \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}(v)}}) \longrightarrow (F_2, \, \pi _2), \;\;\; g_\cdot B(v) \longmapsto [g, \, {\overline{v}}]_{F_2}, \quad g \in G, \end{aligned}$$

is a Poisson embedding.

1.2 Some auxiliary Poisson morphisms

Recall that for any $w \in W$, $B^-wB/B$ is a Poisson submanifold of $(G/B, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})$ (see [24, Theorem 1.5]), and recall that BwB is a Poisson submanifold of $(G, \pi _{\mathrm{st}})$. Recall also from Sect. 2.3 that using the product decomposition ${\overline{w}}N^- {\overline{w}}^{\, 1} = N_wN_w^-=N_w^-N_w$, where again

$$\begin{aligned} N_w = N \cap ({\overline{w}}N^- {\overline{w}}^{\, -1}) \quad \text{ and } \quad N_w^- = N^- \cap ({\overline{w}}N^- {\overline{w}}^{\, -1}), \end{aligned}$$

every $a \in {\overline{w}}N^- {\overline{w}}^{\, -1}$ can be uniquely written as

$$\begin{aligned} a = a_+ a_- = a^\prime _- a^\prime _+ \quad \text{ with } \quad a_+, a^\prime _+ \in N_w, \; a_-, a^\prime _- \in N_w^-. \end{aligned}$$

(A.12)

Lemma A.4

For any $w \in W$, both maps

$$\begin{aligned}&p^w_1: \;\; (wB^-B, \, \pi _{\mathrm{st}}) \longrightarrow (B^-wB/B, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}), \;\; a {\overline{w}}b \longmapsto a_-{\overline{w}}_\cdot B, \quad a \in {\overline{w}}N^- {\overline{w}}^{\, -1}, \; b \in B,\\&p^w_2: \;\; (wB^-B, \, \pi _{\mathrm{st}}) \longrightarrow (BwB, \pi _{\mathrm{st}}), \;\; a{\overline{w}}b \longmapsto a^\prime _+{\overline{w}}b, \quad a \in {\overline{w}}N^- {\overline{w}}^{\, -1}, \; b \in B, \end{aligned}$$

are both Poisson, where $a \in {\overline{w}}N^- {\overline{w}}^{\, -1}$ is decomposed as in (A.12). Furthermore, equip T with the zero Poisson structure. Then the map

$$\begin{aligned} j_w \;\; (wB^-B, \, \pi _{\mathrm{st}}) \longrightarrow (T, 0), \;\; a{\overline{w}}nt \longmapsto t, \quad a \in {\overline{w}}N^- {\overline{w}}^{\, -1}, n \in N, \, t \in N, \end{aligned}$$

is also Poisson.

Proof

Since the projection $(G, \pi _{\mathrm{st}})$ to $(G/B, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})$ is Poisson, to prove $p^w_1$ is Poisson, it suffices to show that

$$\begin{aligned} {\tilde{p}}^w_1: \;\; (wB^-B/B, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}) \longrightarrow (B^-wB/B, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}), \;\; a {\overline{w}}_\cdot B \longmapsto a_-{\overline{w}}_\cdot B, \end{aligned}$$

is Poisson, where again $a \in {\overline{w}}N^- {\overline{w}}^{\, -1}$ is decomposed as in (A.12). For $g,h \in G$, let

$$\begin{aligned}&\sigma _{g}:\;\;\; G/B \longrightarrow G/B, \;\; \;g'_\cdot B \longmapsto gg'_\cdot B, \quad g'\in G,\\&\sigma _{h_\cdot {\scriptscriptstyle B}}:\;\; \;G \longrightarrow G/B, \;\;\; h'\longrightarrow h'h_\cdot B, \quad h' \in G. \end{aligned}$$

Since the left action of $(G, \pi _{\mathrm{st}})$ on $(G/B, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})$ is Poisson, one has

$$\begin{aligned} {\tilde{p}}^w_1(a {\overline{w}}_\cdot B) = {\tilde{p}}^w_1(a_+a_- {\overline{w}}_\cdot B)= ({\widetilde{p}}^w_1\circ \sigma _{a_+}) (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}(a_-{\overline{w}}_\cdot B)) + ({\tilde{p}}^w_1 \circ \sigma _{a_-{\overline{w}}_\cdot {\scriptscriptstyle B}}) (\pi _{\mathrm{st}}(a_+)). \end{aligned}$$

Since $B^-wB/B$ is a Poisson submanifold of $(G/B, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})$, one has

$$\begin{aligned} ({\widetilde{p}}^w_1\circ \sigma _{a_+}) (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}(a_-{\overline{w}}_\cdot B)) =\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}(a_-{\overline{w}}_\cdot B). \end{aligned}$$

Since $N_w$ is a coisotropic subgroup of $(G, \pi _{\mathrm{st}})$, $({\tilde{p}}^w_1 \circ \sigma _{a_-{\overline{w}}_\cdot {\scriptscriptstyle B}}) (\pi _{\mathrm{st}}(a_+))=0$. Thus

$$\begin{aligned} {\tilde{p}}^w_1(a {\overline{w}}_\cdot B) = \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}(a_-{\overline{w}}_\cdot B). \end{aligned}$$

This shows that ${\tilde{p}}^w_1$ is Poisson. Similarly, using the multiplicativity of $\pi _{\mathrm{st}}$ and the fact that $N_w^-$ is a coisotropic subgroup of $(G, \pi _{\mathrm{st}})$ (which can be proved using a similar argument as that in the proof of [41, Lemma 10]), one shows that $p_2^w$ is Poisson.

To show that $j_w$ is Poisson, since that $p_2^w$ is Poisson, it suffices to show that

$$\begin{aligned} j_w^\prime : (BwB, \pi _{\mathrm{st}}) \longrightarrow (T, 0),\;\; j_w^\prime (g' t) = t, \quad g' \in N_w {\overline{w}}N, \, t \in T, \end{aligned}$$

is Poisson. By [41, Lemma 10], both $N_w{\overline{w}}$ and N are coisotropic submanifolds of $(G, \pi _{\mathrm{st}})$. By the multiplicativity of $\pi _{\mathrm{st}}$, $N_w {\overline{w}}N$ is also coisotropic with respect to $\pi _{\mathrm{st}}$. Writing $g \in BwB$ uniquely as $g = g't$, where $g' \in N_w {\overline{w}}N$ and $t \in T$, one has $\pi _{\mathrm{st}}(g) = r_t \pi _{\mathrm{st}}(g')$. Hence $j_w^\prime (\pi _{\mathrm{st}}(g)) = 0$. $\square$

For $w \in W$ and $Q = B(v)$ or N(v), recall from (2.19) the isomorphism

$$\begin{aligned} I^w_{\scriptscriptstyle Q}:&\;\; wB^-B/Q \longrightarrow (B^-wB/B) \times (BwB/Q) \subset (G/B) \times (G/Q), \\&\quad a{\overline{w}}b_\cdot Q \longmapsto (a_-{\overline{w}}_\cdot B, \; a^\prime _+ {\overline{w}}b_\cdot Q), \quad a \in {\overline{w}}N^- {\overline{w}}^{\, -1}, \, b \in B, \end{aligned}$$

where, again, $a \in {\overline{w}}N^- {\overline{w}}^{\, -1}$ is decomposed as in (A.12). Note that $I^w_{\scriptscriptstyle Q}$ is T-equivariant, where T acts on both G/B and G/Q by left translation and on $G/B \times G/Q$ diagonally. For $\xi \in {\mathfrak {h}}= \mathrm{Lie}(T)$, let $\rho _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}(\xi )$ be the vector field on G/Q given by

$$\begin{aligned} \rho _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}(\xi )(g_\cdot Q) = \frac{d}{dt}|_{t = 0} \exp (t\xi )g_\cdot Q, \quad g \in G. \end{aligned}$$

(A.13)

Let again $\{H_q\}_{q=1}^d$ be a basis of ${\mathfrak {h}}$ that is orthonormal with respect to $\langle , \rangle$. Introduce the mixed product Poisson structure $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}} \bowtie _{\mu _0} \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}$ (see Sect. 4.2) on $(G/B) \times (G/Q)$ by

$$\begin{aligned}&\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}} \bowtie _{\mu _0} \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}} = (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}, \; 0) + (0, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}) +\mu _0, \quad \text{ where }\\&\mu _0 = \sum _{q=1}^d (\rho _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}(H_q), \; 0) \wedge (0, \;\rho _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}(H_q)). \end{aligned}$$

It follows from the definition of $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}} \bowtie _{\mu _0} \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}$ that $(B^-wB/B) \times (BwB/Q)$ is a Poisson submanifold of $((G/B) \times (G/Q), \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}} \bowtie _{\mu _0} \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})$.

Lemma A.5

For every $w \in W$, the map

$$\begin{aligned} I_{\scriptscriptstyle Q}^w: \;\; (wB^-B/Q, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}) \longrightarrow ((B^-wB/B) \times (BwB/Q), \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}} \bowtie _{\mu _0} \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})\qquad \quad \end{aligned}$$

(A.14)

is a Poisson isomorphism.

Proof

Since $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}$ is a quotient Poisson structure, $I^w_{\scriptscriptstyle Q}$ is Poisson as long as $I^w_{\{e\}}$ is Poisson. Assume thus $Q = \{e\}$ and note that in this case $G/Q = G$ so $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}} = \pi _{\mathrm{st}}$. Consider the open submanifold $(wB^-B) \times (wB^-B)$ of $G \times G$ and the map

$$\begin{aligned}&D_w: \; (wB^-B) \times (wB^-B) \longrightarrow (B^-wB/B) \times (BwB), \\&\quad (a{\overline{w}}b_1, \; c{\overline{w}}b_2) \longmapsto (a_-{\overline{w}}_\cdot B, \; c^\prime _+{\overline{w}}b_2 ), \end{aligned}$$

where $a, c \in {\overline{w}}N^- {\overline{w}}^{\, -1}$, $b_1, b_2\in B$, and $a = a_+a_-$ and $c = c^\prime _- c^\prime _+$ with $a_+, c^\prime _+ \in N_w$ and $a_-, c^\prime _- \in N_w^-$. We first prove that

$$\begin{aligned}&D_w: \; \left( (wB^-B) \times (wB^-B), \; \Pi _{\mathrm{st}}\right) \nonumber \\&\quad \longrightarrow \left( (B^-wB/B) \times (BwB), \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}} \bowtie _{\mu _0} \pi _{\mathrm{st}}\right) \end{aligned}$$

(A.15)

is Poisson. Indeed, recall that $\Pi _{\mathrm{st}}= (\pi _{\mathrm{st}}, 0) + (0, \pi _{\mathrm{st}}) + \mu _1 + \mu _2$, where $\mu _1$ and $\mu _2$ are respectively given in (A.6) and (A.7). By Lemma A.4, one has

$$\begin{aligned} D_w((\pi _{\mathrm{st}}, 0) + (0, \pi _{\mathrm{st}})) = (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}, \; 0) + (0, \; \pi _{\mathrm{st}}). \end{aligned}$$

By the definition of $D_w$, $D_w(x^L, 0) = 0$ for all $x \in {\mathfrak {b}}=\mathrm{Lie}(B)$. Thus $D_w(\mu _2) = 0$. It also follows from the definition of $D_w$ that for $\alpha \in \Delta ^+$, if $w^{-1}(\alpha ) \in -\Delta ^+$, then $D_w((e_\alpha ^R, 0)) = 0$, and if $w^{-1}(\alpha ) \in \Delta ^+$, then $D_w((0, e_{-\alpha }^R)) = 0$. Consequently,

$$\begin{aligned} D_w(\mu _1) = D_w\left( \sum _{q}^d (H^R_q, 0) \wedge (0, H^R_q)\right) = \mu _0. \end{aligned}$$

This shows that $D_w$ in (A.15) is Poisson. As the diagonal embedding $(wB^-B, \pi _{\mathrm{st}}) \rightarrow \left( (wB^-B) \times (wB^-B), \Pi _{\mathrm{st}}\right)$ is Poisson, we see that $I^w_{\{e\}}$ is Poisson. $\square$

We now turn to the isomorphism $\zeta ^w: B^-wB/B\rightarrow {{\mathcal {O}}}^{w_0w^{-1}}$, for $w \in W$, given in (2.22). Recall also that $t^u = {\overline{u}}^{\, -1} t {\overline{u}}\in T$ for $t \in T$ and $u \in W$.

Lemma A.6

For any $w \in W$, the map

$$\begin{aligned} \zeta ^w: \;(B^-wB/B, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}) \longrightarrow ({{\mathcal {O}}}^{w_0w^{-1}}, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}),\;\; m{\overline{w}}_\cdot B \longmapsto (\overline{w_0w^{-1}}\, m^{-1})_\cdot B, \;\; m \in N_w^-, \end{aligned}$$

is a T-equivariant Poisson isomorphism, where $t \in T$ acts on $B^-wB/B$ by left translation by t and on ${{\mathcal {O}}}^{w_0w^{-1}}$ by left translation by $t^{ww_0}$.

Proof

Let $u = w_0w^{-1}$ so that ${\overline{u}}= \overline{w_0} \,{\overline{w}}^{\, -1}$. It follows from $N_u = {\overline{u}}\,N_w^{-} \,{\overline{u}}^{\, -1}$ that $\zeta ^w$ is a well-defined T-equivariant isomorphism with the T-actions on both sides as described. To show that $\zeta ^w$ is a Poisson isomorphism, consider the two Poisson isomorphisms

$$\begin{aligned}&\zeta _1: \;\; \; (G/B, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}) \longrightarrow (G/B^-, \, \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}}^-}), \;\;\; \zeta _1(g_\cdot B) = g(\overline{w_0}^{\, -1})_\cdot B^-, \quad g \in G,\\&\zeta _2: \;\;\; (G/B^-, \, \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}}^-}) \longrightarrow (B^-\backslash G, \, -\pi _{{{\scriptscriptstyle B}}^-\backslash {\scriptscriptstyle G}}), \;\;\; \zeta _2(g_\cdot B^-) = {B^-}_\cdot g^{-1}, \quad g \in G, \end{aligned}$$

where $\pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}}^-}$ and $\pi _{{{\scriptscriptstyle B}}^-\backslash {\scriptscriptstyle G}}$ respectively denote the Poisson structures on $G/B^-$ and $B^-\backslash G$ that are projections of $\pi _{\mathrm{st}}$ on G. The restriction of the composition $\zeta _2 \circ \zeta _1$ to $(B^-wB/B, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}) \subset (G/B, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})$ gives the Poisson isomorphism

$$\begin{aligned} \zeta _3:&\;\; (B^-wB/B, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})\longrightarrow (B^-\backslash B^-uB^-,\, -\pi _{{{\scriptscriptstyle B}}^-\backslash {\scriptscriptstyle G}}),\\&\;\;\zeta _3(m{\overline{w}}_\cdot B) = {B^-}_\cdot \, \overline{w_0} (m{\overline{w}})^{-1} = {B^-}_\cdot ({\overline{u}}\, m^{-1}), \quad m \in N_w^-. \end{aligned}$$

Note that ${\overline{u}}N_w^- = N {\overline{u}}\cap {\overline{u}}N^-$. By [41, Lemma 14], the map

$$\begin{aligned} \zeta _{{\overline{u}}}: (B^-\backslash B^-uB^-, \, -\pi _{{{\scriptscriptstyle B}}^-\backslash {\scriptscriptstyle G}})\longrightarrow (BuB/B, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}}), \;\; \zeta _{{\overline{u}}}({B^-}_\cdot x) = x_\cdot B, \;\;x \in N{\overline{u}}\cap {\overline{u}}N^-, \end{aligned}$$

is a Poisson isomorphism. As $\zeta ^w = \zeta _{{\overline{u}}} \circ \zeta _3$, we see that $\zeta ^w$ is a Poisson isomorphism. $\square$

For $v, w \in W$, we now relate $(BwB/N(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)})$ and $(BwB/B(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)})$. Let

$$\begin{aligned} K^w_v: BwB/N(v) \longrightarrow (BwB/B(v)) \times T, \;\; n_1{\overline{w}}n_2 t_\cdot N(v) \longmapsto ({n_1{\overline{w}}n_2}_\cdot B(v), \; t),\qquad \end{aligned}$$

(A.16)

where $n_1 \in N_w, \, n_2 \in N_v$, and $t \in T$. It is clear that $K^w_v$ is a T-equivariant isomorphism, where $t_1 \in T$ acts on BwB/N(v) by left translation by $t_1$ and on $(BwB/B(v)) \times T$ by

$$\begin{aligned} t_1 \cdot (g_\cdot B(v), \, t) = (t_1 g_\cdot B(v), \; t_1^wt), \quad t_1, \, t \in T, \, g \in G. \end{aligned}$$

Define the bi-vector field $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)} \bowtie _{\mu ^\prime } 0$ on $(BwB/B(v)) \times T$ by

$$\begin{aligned}&\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)} \bowtie _{\mu ^\prime } 0 = (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)}, \; 0) + \mu ^\prime , \quad \text{ where } \nonumber \\&\mu ^\prime = -\sum _{q=1}^d (\rho _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)}(w(H_q)), \; 0) \wedge (0, \; H_q^R). \end{aligned}$$

(A.17)

Lemma A.7

For any $w, v \in W$,

$$\begin{aligned} K^w_v: \; (BwB/N(v), \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) \longrightarrow ((BwB/B(v)) \times T, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)} \bowtie _{\mu ^\prime } 0) \end{aligned}$$

is a Poisson isomorphism.

Proof

Since the projections

$$\begin{aligned}&(BwB,\; \pi _{\mathrm{st}}) \longrightarrow (BwB/N(v), \;\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}),\\&(BwB/T, \;\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle T}}) \longrightarrow (BwB/B(v), \;\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)}) \end{aligned}$$

are T-equivariant and Poisson, it is enough to prove Lemma A.7 for $v = {w_0}$, i.e., to prove that

$$\begin{aligned} K := K^w_{w_0}:\;(BwB, \,\pi _{\mathrm{st}}) \longrightarrow ((BwB/T) \times T, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle T}} \bowtie _{\mu ^\prime } 0) \end{aligned}$$

is Poisson. Consider again $G \times G$ with the multiplicative Poisson structure $\Pi _{\mathrm{st}}$ from (A.4). By (A.5), $(BwB) \times G$ is a Poisson submanifold of $(G \times G, \Pi _{\mathrm{st}})$. Define

$$\begin{aligned}&K^\prime : \; BwB \times G \longrightarrow (G/T) \times T, \;\;(g' t, \; g) \longmapsto (g_\cdot T, \; t),\quad \\&g' \in N{\overline{w}}N, \, g \in G, \, t \in T. \end{aligned}$$

Since K is the composition of $K^\prime$ with the diagonal Poisson embedding $(BwB, \pi _{\mathrm{st}})\hookrightarrow (BwB \times G, \Pi _{\mathrm{st}})$, K is Poisson once we prove that

$$\begin{aligned} K^\prime : (BwB \times G, \;\Pi _{\mathrm{st}}) \longrightarrow ((G/T) \times T, \; (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle T}}, 0) + \mu ^\prime ) \end{aligned}$$

is Poisson. Recall again that $\Pi _{\mathrm{st}}= (\pi _{\mathrm{st}}, 0) + (0, \pi _{\mathrm{st}}) + \mu _1 + \mu _2$, with $\mu _1$ and $\mu _2$ respectively given in (A.6) and (A.7). By Lemma A.4, $K^\prime (\pi _{\mathrm{st}}, 0)=0$. By the definition of $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle T}}$, $K^\prime (0, \pi _{\mathrm{st}}) = (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle T}}, 0)$. Thus

$$\begin{aligned} K^\prime ((\pi _{\mathrm{st}}, 0) + (0, \pi _{\mathrm{st}})) = (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle T}}, 0). \end{aligned}$$

It follows from the definitions that $K^\prime (e_\alpha ^L, 0) = K^\prime (e_\alpha ^R, 0) = 0$ for all $\alpha \in \Delta ^+$ and that $K^\prime (0, x^L) = 0$ for $x \in {\mathfrak {h}}$. Furthermore, it follows from the definition of $K^\prime$ that

$$\begin{aligned} K^\prime (x^R, 0) = (0, \, (w^{-1}(x))^R) \quad \text{ and } \quad K^\prime (0, x^R) = (\rho _{{\scriptscriptstyle G}/{\scriptscriptstyle T}}(x), \, 0), \quad x \in {\mathfrak {h}}. \end{aligned}$$

The fact that $K^\prime$ is Poisson now follows from

$$\begin{aligned} K^\prime (\mu _1+\mu _2)&= K^\prime \left( \sum _{q=1}^d(H_q^R, \,0) \wedge (0, \, H_q^R)\right) \\&=\sum _{q=1}^d (0, \, (w^{-1}(H_q))^R) \wedge (\rho _{{\scriptscriptstyle G}/{\scriptscriptstyle T}}(H_q), \, 0) = \mu ^\prime . \end{aligned}$$

$\square$

For $v, w \in W$, recall from (2.20) and (2.21) the isomorphisms

$$\begin{aligned}&\zeta ^{(w, v)}_{{\scriptscriptstyle B}(v)}: \;\; BwB/B(v) \longrightarrow {{\mathcal {O}}}^{(w, v)}, \;\;\; {n_1{\overline{w}}n_2}_\cdot B(v) \longmapsto [n_1{\overline{w}}, \; n_2{\overline{v}}]_{F_2},\\&{\zeta }^{(w, v)}_{{\scriptscriptstyle N}(v)}: \;\; BwB/N(v) \longrightarrow {{\mathcal {O}}}^{(w, v)} \times T, \;\;\; n_1{\overline{w}}n_2 t_\cdot N(v) \longmapsto ([n_1{\overline{w}}, \; n_2{\overline{v}}]_{F_2},\, t), \end{aligned}$$

where $n_1 \in N_w, \, n_2 \in N_v$ and $t \in T$. Note that $\zeta ^{(w, v)}_{{\scriptscriptstyle B}(v)}$ and ${\zeta }^{(w, v)}_{{\scriptscriptstyle N}(v)}$ are T-equivariant, where $t_1 \in T$ acts on BwB/B(v) and BwB/N(v) by left translation by $t_1$ and on ${{\mathcal {O}}}^{(w, v)} \subset F_2$ and on ${{\mathcal {O}}}^{(w, v)} \times T$ respectively by (see (2.10))

$$\begin{aligned}&t_1 \cdot [n_1{\overline{w}}, \, n_2{\overline{v}}]_{{\scriptscriptstyle F}_2} = [t_1n_1{\overline{w}}, \, n_2{\overline{v}}]_{{\scriptscriptstyle F}_2}, \\&t_1 \cdot ([n_1{\overline{w}}, \, n_2{\overline{v}}]_{{\scriptscriptstyle F}_2}, \, t) = ([t_1n_1{\overline{w}}, \, n_2{\overline{v}}]_{{\scriptscriptstyle F}_2}, \, t_1^wt), \quad n_1 \in N_w, \, n_2 \in N_v, \, t \in T. \end{aligned}$$

Equip ${{\mathcal {O}}}^{(w, v)}$ with the standard Poisson structure $\pi _{2}$ given in (4.10), and let $\mu ^{\prime \prime }$ be the bi-vector field on ${{\mathcal {O}}}^{(w, v)} \times T$ by

$$\begin{aligned} \mu ^{\prime \prime } = -\sum _{q=1}^d (\rho _2(w(H_q)), \, 0) \wedge (0, \, H_q^R), \end{aligned}$$

(A.18)

where $\rho _2$ is defined in (5.1).

Lemma A.8

The maps

$$\begin{aligned}&\zeta ^{(w, v)}_{{\scriptscriptstyle B}(v)}: \;\; (BwB/B(v), \;\pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}(v)}}) \longrightarrow ({{\mathcal {O}}}^{(w, v)}, \;\pi _{2}),\\&{\zeta }^{(w, v)}_{{\scriptscriptstyle N}(v)}: \;\; (BwB/N(v), \;\pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle N}(v)}}) \longrightarrow ({{\mathcal {O}}}^{(w, v)} \times T, \;(\pi _{2}, \, 0) + \mu ^{\prime \prime }) \end{aligned}$$

are Poisson isomorphisms.

Proof

As BwB/B(v) is a Poisson submanifold of $(G/B(v), \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}(v)}})$, by Lemma A.3, one has the T-equivariant Poisson embedding

$$\begin{aligned} (BwB/B(v), \; \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}(v)}}) \longrightarrow (F_2, \, \pi _2), \;\;n{\overline{w}}b_\cdot B(v)\longmapsto [n{\overline{w}}b, \, {\overline{v}}]_{{\scriptscriptstyle F}_2} = [n{\overline{w}}, \, b{\overline{v}}]_{{\scriptscriptstyle F}_2}, \end{aligned}$$

where $n \in N_w$ and $b \in B$. Since the image of the above embedding is precisely ${{\mathcal {O}}}^{(w, v)}$, we see that the $\zeta ^{(w, v)}_{{\scriptscriptstyle B}(v)}$ is a Poisson isomorphism. The fact that ${\zeta }^{(w, v)}_{{\scriptscriptstyle N}(v)} = (\zeta ^{(w, v)}_{{\scriptscriptstyle B}(v)} \times \mathrm{Id}) \circ K^w_v$ is a Poisson isomorphism follows directly from Lemma A.7. $\square$

1.3 Proof of Theorem 5.3

Let again $w, v \in W$, and let the notation be as in the statement of Theorem 5.3. First let $Q = B(v)$. By Lemmas A.5, A.6, and A.8, one has the Poisson isomorphisms

(A.19)

Since $J^w_{{\scriptscriptstyle B}(v)} = (\zeta ^w \times \zeta ^{(w, v)}_{{\scriptscriptstyle B}(v)}) \circ I^w_{{\scriptscriptstyle B}(v)}$, one sees that

$$\begin{aligned} J^w_{{\scriptscriptstyle B}(v)}: \;\; (wB^-B/B(v), \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)}) \longrightarrow \left( {{\mathcal {O}}}^{w_0w^{-1}} \times {{\mathcal {O}}}^{(w, v)}, \; \pi _{1, 2}\right) \end{aligned}$$

is Poisson. To see that $J^w_{{\scriptscriptstyle N}(v)}$ is Poisson, note that

$$\begin{aligned} J^w_{{\scriptscriptstyle N}(v)} = \left( \mathrm{Id} \times {\zeta }^{(w, v)}_{{\scriptscriptstyle N}(v)}\right) \circ (\zeta ^w \times \mathrm{Id}) \circ I^w_{{\scriptscriptstyle N}(v)}. \end{aligned}$$

By Lemmas A.5 and A.6,

$$\begin{aligned} \left( (\zeta ^w \times \mathrm{Id})\circ I^w_{{\scriptscriptstyle N}(v)}\right) (\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) = (\pi _1, \, 0) + (0, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) +\mu _0^\prime , \end{aligned}$$

where $\mu _0^\prime = \sum _{q=1}^d (\rho _1(w_0w^{-1}(H_q)), \, 0) \wedge (0, \, \rho _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}(H_q))$. By Lemma A.8 and the T-equivariance of ${\zeta }^{(w, v)}_{{\scriptscriptstyle N}(v)}$, one has

$$\begin{aligned}&(\mathrm{Id} \times {\zeta }^{(w, v)}_{{\scriptscriptstyle N}(v)})((\pi _1, \, 0) + (0, \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) +\mu _0^\prime ) = (\pi _1, 0, 0) +(0, \, \pi _2, \, 0) + (0, \, \mu ^{\prime \prime }) \\&\quad + \sum _{q=1}^d \left( (\rho _1(w_0w^{-1}(H_q)), \, 0, \, 0) \wedge ((0, \, \rho _2(H_q), \, 0) + (0, \, 0, \, (w^{-1}(H_q))^R))\right) \\&\quad =(\pi _1, 0, 0) + (0, \, \pi _2, \, 0) + \mu _{23} + \mu _{12} + \mu _{13} = \pi _{1, 2, 0}. \end{aligned}$$

It follows that $J^w_{{\scriptscriptstyle N}(v)}: (wB^-B/N(v), \, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) \rightarrow ({{\mathcal {O}}}^{w_0w^{-1}} \times {{\mathcal {O}}}^{(w, v)} \times T, \, \pi _{1,2,0})$ is Poisson. This finishes the proof of Theorem 5.3.

1.4 T-leaves of $(G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})$

For any Poisson variety $(X, {\pi _{\scriptscriptstyle X}})$ with an action by an algebraic ${{\mathbb {C}}}$-torus ${{\mathbb {T}}}$ via Poisson isomorphisms, define (see [40]) the ${{\mathbb {T}}}$-leaf of ${\pi _{\scriptscriptstyle X}}$ through $x \in X$ to be ${{\mathbb {T}}}\Sigma _x = \bigcup _{t \in {{\mathbb {T}}}} t\Sigma _x$, where $\Sigma _x$ is the symplectic leaf of ${\pi _{\scriptscriptstyle X}}$ through x. For the T-Poisson variety $(G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})$, where $Q = B(v)$ or N(v) for $v \in W$ and T acts on G/Q by left translation, we now determine the T-leaves of $(G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})$.

Let $\varpi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}}: G \rightarrow G/Q$ be the projection. Recall the monoidal product $*$ on W determined by $w *s_\alpha = ws_\alpha$ if $l(ws_\alpha ) =l (w) + 1$ and $w *s_\alpha = w$ if $l(ws_\alpha ) = l(w) - 1$.

Theorem A.9

For $v \in W$ and $Q = B(v)$ or N(v), the T-leaves of $(G/Q, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}})$ are precisely the subvarieties

$$\begin{aligned} L_{w, y}^{{\scriptscriptstyle G}/{\scriptscriptstyle Q}} \, {\mathop {=}\limits ^{{\mathrm{def}}}} \,\varpi _{{\scriptscriptstyle G}/{\scriptscriptstyle Q}} \left( (BwB) \cap B^-yB {\overline{v}}^{\, -1}\right) \subset G/Q, \end{aligned}$$

where $y, w \in W$ and $y \le w *v$.

Proof

Consider first the case of $Q = B(v)$ and recall from Lemma A.3 the T-equivariant Poisson embedding

$$\begin{aligned} E_v:\;\; (G/B(v), \; \pi _{{\scriptscriptstyle G}/{{\scriptscriptstyle B}(v)}}) \longrightarrow (F_2, \, \pi _2), \;\;\; g_\cdot B(v) \longmapsto [g, \, {\overline{v}}]_{F_2}, \quad g \in G, \end{aligned}$$

where T acts on $F_2$ by (2.10). It follows from the Bruhat decomposition $G = \bigsqcup _{w \in W} BwB$ that the image of $E_v$ is given by

$$\begin{aligned} E_v(G/B(v)) = \bigsqcup _{w \in W} {{\mathcal {O}}}^{(w, v)} \subset F_2. \end{aligned}$$

The T-leaves of $(F_r, \pi _r)$, for any $r \ge 1$, are determined in [40, Theorem 1.1]. For the case of $r = 2$ at hand, let

$$\begin{aligned} \mu _{{\scriptscriptstyle F}_2}: \;\; F_2 \longrightarrow G/B, \;\; [g_1, g_2]_{{\scriptscriptstyle F}_2} \longmapsto {g_1g_2}_\cdot B. \end{aligned}$$

By [40, Theorem 1.1], the T-leaves of $(F_2, \pi _2)$ are precisely the intersections

$$\begin{aligned} R^{(w, x)}_y \, {\mathop {=}\limits ^\mathrm{def}} \, {{\mathcal {O}}}^{(w, x)} \cap \mu _{{\scriptscriptstyle F}_2}^{-1}(B^-yB/B), \end{aligned}$$

where $w, x, y \in W$ and $y \le w*x$. Thus, for each $w \in W$, ${{\mathcal {O}}}^{(w, v)}\subset F_2$ is a union of all the T-leaves $R^{(w, v)}_y$ with $y \le w *v$. It is easy to see that $L_{w, y}^{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)} = E_v^{-1}(R^{(w, v)}_y)$ for all $w, y \in W$ with $y \le w*v$. Thus the $L_{w, y}^{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)}$’s are precisely all the T-leaves of $(G/B(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)})$.

Consider now $Q = N(v)$ and the decomposition $G/N(v) = \bigsqcup _{w \in W} BwB/N(v)$, where each BwB/N(v) is a T-invariant Poisson submanifold with respect to $\pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}$. Let $w \in W$ and recall from Lemma A.7 the T-equivariant Poisson isomorphism

$$\begin{aligned} K^w_v: \; (BwB/N(v), \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) \longrightarrow ((BwB/B(v)) \times T, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)} \bowtie _{\mu ^\prime } 0), \end{aligned}$$

where $t_1 \in T$ acts on $(BwB/B(v)) \times T$ by

$$\begin{aligned} t_1 \cdot (g_\cdot B(v), \, t) = (t_1 g_\cdot B(v), \; t_1^wt), \quad t_1, \, t \in T, \, g \in G. \end{aligned}$$

By [38, Lemma 2.23], the T-leaves of $((BwB/B(v)) \times T, \; \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)} \bowtie _{\mu ^\prime } 0)$ are precisely of the form $L \times T$, where L is a T-leaf of $(BwB/B(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}(v)})$. Applying the T-equivariant Poisson isomorphism $K^w_v$, one sees that the T-leaves of $(BwB/N(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)})$ are precisely $L_{w, y}^{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}$, where $y \in W$ and $y \le w *v$. It follows that $L_{w, y}^{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}$, where $w, y \in W$ and $y \le w *v$, are all the T-leaves of $(G/N(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)})$. $\square$

Example A.10

When $v = w_0$, so that $G/N(v) = G$, one has $w*w_0 = w_0$ for every $w \in W$, so the condition $y \le w*w_0$ is satisfied for every $y \in W$, and one has $B^-yB\overline{w_0}^{-1} = B^-yw_0B^-$. In this case, Theorem A.9 recovers the well-know result [32, 33, 36] that the T-leaves (for the T-action on G by left translation) of $(G, \pi _{\mathrm{st}})$ are precisely all the double Bruhat cells $G^{w, u} = BwB \cap B^-uB^-$, where $w,u \in W$.

When $v = e$, so that $G/B(v) = G/B$, Theorem A.9 recovers the well-know result from [24] that the T-leaves of $(G/B, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle B}})$ are precisely the open Richardson varieties $(BwB/B) \cap (B^-yB/B)$, where $w, y \in W$ and $y \le w$. $\square$

The next examples shows that for any $w, v \in W$, the double Bruhat cell $G^{w, v^{-1}}$, as a T-leaf of $(G, \pi _{\mathrm{st}})$, can also be embedded into G/N(v) as a T-leaf of $(G/N(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)})$.

Example A.11

For an arbitrary $v \in W$, consider the T-leaf

$$\begin{aligned} L_{w, e}^{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)} =\varpi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)} \left( (BwB) \cap B^-B {\overline{v}}^{\, -1}\right) \end{aligned}$$

of $(G/N(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)})$. Recall from Lemma 3.7 the embedding

$$\begin{aligned} \delta _v: \;\; B^-v^{-1}B^- \longrightarrow G/N(v), \;\; g \longmapsto g_\cdot N(v). \end{aligned}$$

For $w \in W$, denote the restriction of $\delta _v$ to $G^{w, v^{-1}} = BwB \cap B^-v^{-1}B^-$ by

$$\begin{aligned} \delta _{w, v} = \delta _v|_{{\scriptscriptstyle G}^{w, v^{-1}}}:\;\;\; G^{w, v^{-1}} \longrightarrow G/N(v). \end{aligned}$$

(A.20)

It then follows that the image of $\delta _{w, v}$ is precisely the T-leaf $L_{w, e}^{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}$ of $(G/N(v), \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)})$. As $G^{w, v^{-1}}$ is a T-leaf of $(G, \pi _{\mathrm{st}})$, we conclude that

$$\begin{aligned} \delta _{w, v}:\;\; (G^{w, v^{-1}}, \pi _{\mathrm{st}}) {\mathop {\longrightarrow }\limits ^{\sim }} (L_{w, e}^{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}, \pi _{{\scriptscriptstyle G}/{\scriptscriptstyle N}(v)}) \end{aligned}$$

is a Poisson isomorphism of T-leaves. $\square$